Question Number 86963 by abdomathmax last updated on 01/Apr/20 $${calculate}\:\:{I}\:=\int\:\:{cos}^{\mathrm{4}} {x}\:{sh}^{\mathrm{2}} {x}\:{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 152492 by nadovic last updated on 28/Aug/21 $$\:\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{upwards}\:\mathrm{with} \\ $$$$\:\mathrm{a}\:\mathrm{velocity}\:\mathrm{of}\:\:\mathrm{96}{ms}^{−\mathrm{1}} .\:\mathrm{In}\:\mathrm{addition}\:\mathrm{to} \\ $$$$\:\mathrm{being}\:\mathrm{subject}\:\mathrm{to}\:\mathrm{gravity},\:\mathrm{it}\:\mathrm{is}\:\mathrm{acted}\:\mathrm{on} \\ $$$$\:\mathrm{by}\:\mathrm{a}\:\mathrm{retardation}\:\mathrm{of}\:\mathrm{16}{t},\:\mathrm{where}\:{t}\:\mathrm{is}\:\mathrm{the} \\ $$$$\:\mathrm{time}\:\mathrm{from}\:\mathrm{the}\:\mathrm{start}\:\mathrm{of}\:\mathrm{the}\:\mathrm{motion}. \\ $$$$\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{height}\:\mathrm{attained} \\ $$$$\:\mathrm{by}\:\mathrm{the}\:\mathrm{particle}? \\ $$…
Question Number 86957 by abdomathmax last updated on 01/Apr/20 $${find}\:\int\:\:{arctan}\left(\frac{\mathrm{1}−{u}}{\mathrm{1}+{u}}\right){du} \\ $$ Commented by Ar Brandon last updated on 01/Apr/20 $${I}=\int{arctan}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right){dx} \\ $$$${Let}\:{u}={arctan}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)\:\Leftrightarrow\:{u}={arctan}\left({t}\right)\:\:{with}\:\:{t}=\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}} \\ $$$$\Rightarrow\frac{{du}}{{dx}}=\frac{{du}}{{dt}}×\frac{{dt}}{{du}}…
Question Number 152494 by mnjuly1970 last updated on 28/Aug/21 $$ \\ $$$$\:\:\:\:{prove}\:{that}\::: \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\Omega\::=\:\int_{\mathrm{0}\:} ^{\:\infty} \frac{\:\:{e}^{\:−{x}} .\mathrm{ln}\:\left(\frac{\:\mathrm{1}}{\:{x}}\:\right)\:{sin}\:\left(\:{x}\:\right)}{{x}\:}\:{dx}\:=\:\frac{\:\pi}{\:\mathrm{8}}\:\left(\:\mathrm{2}\:\gamma\:+\mathrm{ln}\:\left(\mathrm{2}\:\right)\:\right)\:…\blacksquare\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:{m}.{n} \\ $$$$ \\ $$…
Question Number 86956 by abdomathmax last updated on 01/Apr/20 $${find}\:\int\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){arctan}\left(\mathrm{2}{x}\right){dx} \\ $$ Commented by Ar Brandon last updated on 01/Apr/20 $${Let}\:{u}={arctan}\left(\mathrm{2}{x}\right)\:\Rightarrow\:{du}=\mathrm{2}\centerdot\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{2}{x}\right)^{\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:{dv}=\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}}…
Question Number 86955 by abdomathmax last updated on 01/Apr/20 $${find}\:\int\:\:\frac{{arctan}\left({x}\right)}{{x}}{dx} \\ $$ Commented by Ar Brandon last updated on 01/Apr/20 $${Let}\:{f}\left({a}\right)=\int\frac{{arctan}\left({ax}\right)}{{x}}{dx} \\ $$$$\Rightarrow{f}\:'\:\left({a}\right)=\int\frac{{x}}{{x}}\centerdot\frac{\mathrm{1}}{\mathrm{1}+\left({ax}\right)^{\mathrm{2}} }{dx}=\int\frac{\mathrm{1}}{\mathrm{1}+\left({ax}\right)^{\mathrm{2}} }{dx}…
Question Number 152472 by talminator2856791 last updated on 28/Aug/21 $$\: \\ $$$$\:\:\:\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\:=\:\mathrm{exp}\left(−\:\frac{\mathrm{3}\gamma}{\mathrm{4}}\:+\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {f}\left({x}\right){dx}\right) \\ $$$$\: \\ $$$$\:\:\:\:\mathrm{find}\:{f}\left({x}\right) \\ $$$$\: \\ $$ Answered by mindispower…
Question Number 152468 by talminator2856791 last updated on 28/Aug/21 $$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{−\infty} ^{\:\infty} \:\frac{\mathrm{ln}\left(\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}\right)}{\:\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}\:\:{dx} \\ $$$$\: \\ $$ Answered by mindispower last updated…
Question Number 21390 by Joel577 last updated on 22/Sep/17 $$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\:\frac{{x}^{\mathrm{7}} \:−\:\mathrm{1}}{\mathrm{ln}\:{x}}\:{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 86914 by Chi Mes Try last updated on 01/Apr/20 Terms of Service Privacy Policy Contact: info@tinkutara.com